Both Jupiter and Saturn have rocky cores. Is there such of a thing as a gaseous planet without a core? And would a planet without a core have gravity?
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$\begingroup$ Just to answer one part of your question, the second part: yes, that theoretical planet would have identical gravity; regarding gravity it makes absolutely no difference what the thing is made of. $\endgroup$– FattieCommented May 28, 2021 at 12:38
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2 Answers
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The gravitational force on a small mass on the outside of a planet is always the Newtonian $$F_{G}=-\frac{GM}{r^2},$$ so any planet, and particularly, any mass in the universe produces a gravitational field acting on everything else. So if, for example, the mass is $M=2\times 10^{27}\rm kg$ (i.e. one Jovian mass), then the gravity field outside the planet will always be the same (apart from Tides, higher order moments), no matter whether the mass is in Hydrogen or solids.
For the gas giants Jupiter and Saturn in our solar system, the mass in heavy refractories (i.e. everything heavier than Helium) is about $M_{\rm ref}\approx 15-20 \rm m_{\oplus}$, where $\rm m_{\oplus}$ is an Earth mass.
The rest of $M$ is hydrogen/helium. For Jupiter this is 300 $\rm m_{\oplus}$, Saturn about $75 \rm m_{\oplus}$.
This is a relatively large number of refractories in those gas giants, compared to solar composition, which is why we think that they have been formed via core-accretion, see Pollack (1996).
However there is another idea of how to form gas giants out there, which is that of gravitational disc instability, see Boss (2002). This idea posits that at very massive protostellar discs, which form planets, can become unstable and fragment into large clumps, which form gas giants directly. Those disc instability giant planets would have solar metallicity, i.e. a Jupiter-mass planet would have a refractory mass of only $M_{\rm ref} \approx 3 \rm m_{\oplus}$.
Those refractories would presumably sink to the planetary center and form a small core. Exoplanets that were found at large semi-major axis distances (hundreds of AU, compared to the Jovian 5 AU) from their stars, such as YSES 2b, are candidates for such disc instability models, and hence would host such small core. But that is as small of a core as it gets, you cannot have a core much much less massive than this.
answered May 27, 2021 at 18:05
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1$\begingroup$ The very first planets to form in the early universe before there were much metals would presumably have much less massive cores. $\endgroup$– snoCommented May 27, 2021 at 19:23
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$\begingroup$ @sno What planets are you talking about? Very low mass rocky ones? Then sure, the occurrence of rocky exoplanets seems to be fairly uniform throughout metallicity space for rockies. Gas giants however need metallicities of $[Fe/H]>-1$ and stellar masses of $M_{\star}<1.1 M_{\odot}$ to form in significant numbers (Adibekyan 2019, Heavy metal rules). So the lowest you can go with a core mass would be a disc instability, $[Fe/H]=-1$ protostellar cloud that forms e.g. Jupiters, with $0.3 m_{\oplus}$ cores. Not lower than that, as stated in my answer. $\endgroup$ Commented May 27, 2021 at 20:30
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$\begingroup$ @sno Furthermore apart from the data, both models, core accretion and disc instability have metallicity requirements to form gas giants. Core accretion needs a minimum amount of metals to cool their envelopes via molecular bands in order to reach runaway gas accretion. Disc instability needs a minimum amount of metals to cool in order to reach the Toomre-criterion of instability. Both minimum requirements work with the found limit of $[Fe/H]\approx 0.1$, so you'd expect no gas giants and hence no cores in the early universe. $\endgroup$ Commented May 27, 2021 at 20:33
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$\begingroup$ What about sub-brown dwarfs forming with no metals and then being captured into orbit around a star to become a planet, or do sub-brown dwarfs also have a minimum metallicity to form? $\endgroup$– snoCommented May 27, 2021 at 20:50
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$\begingroup$ @sno: Brown dwarves are actually exceedingly rare, as they form the low-mass end of the star formation IMF (which depends on metallicity) and are separated from the high-mass end of giant planets (the so-called "Brown Dwarf desert" ui.adsabs.harvard.edu/abs/2006ApJ...640.1051G/abstract). The lower mass of BDs will only increase as metallicity decreases as the Jeans cooling required for clumping follows a similar logic as the other processes (academic.oup.com/mnras/article/363/2/363/1123406). $\endgroup$ Commented May 27, 2021 at 21:07
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Anything with mass has gravity, so yes, such a planet would have gravity.
However, gases tend to disperse in their surrounding environment, so you’d need a very massive gas cloud to collapse into such a planet for gases not to disperse. This raises the question of the pressure at the centre of this planet; it would be high enough to turn the gas at least into a liquid, if not a solid. Another possibility is that the gas at the core would turn into a plasma (such as in the centre of the Sun) because of the heat—a plasma is basically a hot gas stripped of some of its electrons.
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$\begingroup$ This begs the question, do we know the core in Jupiter and Saturn is rocky instead of a solid gas? My understanding is that we know about the core of both planets only because of the measure of their oblateness. $\endgroup$ Commented May 27, 2021 at 17:45
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1$\begingroup$ @Bookaholic: That is not true. The existence of a core in the gas giants has long been proposed to be true due to studies of the equation of state of Hydrogen/Helium mixtures at high pressures. If you take a Jupiter's mass of pure H/He at Jupiters luminosity, you cannot fit this into a ball of one Jupiter-radius, the ball gets too massive. Only by including a ~15$m_{oplus}$ solid mass inside can you get a consistent model Jupiter. Furthermore, in recent years fuzzy cores (or metallicity gradients to be more precise) inside J and S have been measured via JUNO and Cassini. $\endgroup$ Commented May 27, 2021 at 18:34
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$\begingroup$ Under which circumstances does a plasma form? $\endgroup$– 2080Commented May 27, 2021 at 18:46
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$\begingroup$ @AtmosphericPrisonEscape Can you define $m_{oplus}$? $\endgroup$ Commented May 28, 2021 at 0:12
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1$\begingroup$ @AlexHajnal, sorry a typo. $\rm m_{oplus}=m_{\oplus} = 5.9\times 10^{27} g$. $\endgroup$ Commented May 28, 2021 at 0:35